Question

If 16.9 kilograms of Al2O3(s), 51.4 kilograms of NaOH(l), and 51.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

Answer 1

Production of cryolite reaction is: Al₂O₃ + 6 NaOH + 12 HF → 2 Na₃AlF₆ + 9 H₂O
from the reaction it is clear that, 101.9 g of Al₂O₃ reacts with 6 x 40 g of NaOH. Then the amount of NaOH need to react with 16.9 kg of Al₂O₃ can be, (16900 g x 6 x 40 g)/101.9g = 39.8 kg of NaOH.

101.9 g of Al₂O₃ reacts with 12 x 20 g of HF. Then the amount of HF need to react with 16.9 kg of Al₂O₃ can be, (16900 g x 12 x 20 g)/101.9g = 39.8 kg of HF.

So Al₂O₃ is the limiting reactant, and NaOH and HF are in excess.

101.9 g of Al₂O₃ produces, 2 x 209.9 g of Na₃AlF₆. Therefore the amount of Na₃AlF₆ produced from 16.9 kg of Al₂O₃ is: (16900 g x 2 x 209.9 g)/101.9g = 69.6 kg of Na₃AlF₆.